In: Math
We have to test the hypothesis that
Ho : P = 0.67 against H1: P # 0.67 ( two-tailed test)
Given : n = 598
X: Observed number of successes = 419
p : sample proportion = 419/ 598 = 0.7126
Sampling distribution of proportion normal
p ~ N ( P, P(1-P) /n)
E(p) = P = 0.67 and SD(P) = sqrt( P * (1-P) /n) = sqrt ( 0.67 *0.33 /598) = 0.0192
The value of test statistic is
z = (p- E(p)) / SD(p) ~ N(0,1)
z = ( 0.7126 - 0.67 ) / 0.0192
z = 2.2187
Value of test statistic z = 2.2187
Since the test is two tailed and value of test statistic is 2.2187, p-value is obtained by
p-value = 2* P ( Z >2.2187)
from normal probability table
P ( Z > 2.2187) = 0.0133
p-value = 0.0266
Alpha: level of significance = 0.01
Decision: since p-value is greater than level of significance alpha, we failed to reject Ho.
Conclusion : There is not sufficient evidence support to claim the population proportion differs from 0.67.